dv/dt = cos(t³) - 3t³sin(t³)

Integrate both sides,

v = ∫ [ (1)cos(t³) + (t)(3t²)(-sin(t³)) ] dt

v = ∫ (d/dt ( tcos(t³) ) dt

v = tcos(t³) + c , c is a constant

when v(0) = 0,

0 = 0(cos(0³)) + c

c = 0

So v = tcos(t³)

how did you integrate these terms? ==> (t)(3t²)(-sin(t³))

Alternatively,

dv/dt - cos(t³) = -3t³sin(t³)

Integrate both sides with respect to t,

v - ∫cos(t³) dt = ∫ -3t³sin(t³) dt

v - ∫cos(t³) dt = ∫ ( t(-3t²sin(t³)) ) dt



Integrating t(-3t²sin(t³)) by parts,

(Formula : ∫(u dv/dt) dt = uv - ∫(v du/dt) dt

We are letting u = t and dv/dt = -3t²sin(t³)


v - ∫cos(t³) dt = tcos(t³) - ∫ (cos(t³)(1)) dt

v = tcos(t³) + ∫cos(t³) dt - ∫cos(t³) dt

v = tcos(t³) + C , C is a constant


Notice ∫cos(t³) gets eliminated.

And that is why we let dv/dt = -3t²sin(t³) as it is directly integrable to cos(t³). Then ∫cos(t³) can be eliminated.


when v(0) = 0,

0 = (0)cos(0³) + C
C = 0

So v = tcos(t³)

In the first solution, when you differentiate tcos(t³) with respect to t,

d/dt (tcos(t³))

= d/dt (t) × cos(t³) + t × d/dt (cos(t³))
(Product rule)

= 1 × cos(t³) + t × (3t²(-sin(t³)))
(Apply chain rule)

= cos(t³) - 3t³sin(t³)

So by rearranging the equation given in the question, you'll realise it's directly integrable back to tcos(t³). So we are working backwards here.


So this question is actually directly solvable but since the question has given you a hint to use integration by parts, I've written the 2nd solution showing how to do so.

how did you identify that -3t²sin(t³) can be directly integrated to cos(t³)?
I would be stuck trying to integrate 3t³sin(t³) by parts

Firstly, the cos(t³) isn't directly integrable. So we'll need to try to manipulate the other terms to get rid of it.

The question expects you to deduce that you'll need to use -3t³sin(t³) to do so. The setter of this question has told you to integrate it by parts and also informed you that the term that is unresolvable (which is the cos(t³) would be cancelled. (In exam you'll have to figure this out on your own!)



Now, most would be stuck trying to integrate -3t³sin(t³) by parts. To integrate by parts , we have to determine what are the two factors to be dealt with. But we aren't sure how.

Most students will try to treat -3t³ as one factor and sin(t³) as the other factor as it seems very intuitive to group it like this. But they will get stuck later and this wastes time in exam. So we need to determine the correct factors before proceeding to integrate.

So we need to look at the formula.
∫(u dv/dt) dt = uv - ∫(v du/dt) dt

dv/dt has to be integrated while u has to be differentiated. Since the right hand side
of the equation has an integral, we want to make ∫(v du/dt) dt = ∫(cos(t³)) dt such that we can cancel out the other ∫cos(t³) dt and get rid of them both.

So here we try to differentiate cos(t³) to see if it gives us any leads/clues. Since differentiating it gives -3t²sin(t³), by letting dv/dt = -3t²sin(t³), integrating back gives us v = cos(t³)

Next,-3t³sin(t³) = t(-3t²sin(t³)). By letting dv/dt = -3t²sin(t³), then u = t and thus du/dt = 1. This fits our requirement perfectly since this makes v du/dt = cos(t³).



In summary what I'm doing here is :

① Realising that cos(t³) isn't directly integrable and deducing it has to be gotten rid of.

② Realising that I have to manipulate -3t³sin(t³) to do so, via integration by parts.

③ Deducing that I can make ∫(v du/dt) dt = ∫(cos(t³)) dt to get rid of the other ∫(cos(t³)) dt.

④ Testing out which factors work before actually proceeding to integrate. I differentiate cos(t³) , then decide what I can set as dv/dt and what can u be.

I have seen the question and J has done a good job of answering the question (I myself am rusty at these).

Very good question coming from you.

Now, let us observe the term -3t2 sin (t3) carefully.

The term outside the sin, which is 3t2 (not counting the - sign) is actually the derivative of t3 (which is inside the sin). This suggests a reverse chain rule process.

Now, we are halfway there after identifying that 3t2 is the derivative of the inside bracker t3 which is the chain. We are now left with finding the integral of sin, which is negative cos.

So, integrating 3t2 sin (t3) gives you - cos (t3). You can re-differentiate this expression - cos (t3) to get back the derivative.

Therefore, integrating -3t2 sin (t3) gives you cos (t3).

In short, as J has noted, sin (t3) (and cos (t3) as well) cannot be integrated immediately, but once you notice that differentiating either of those will result in a "3t2" chain, then we introduce the 3t2 accordingly.

So here, 3t3 sin (t3) is to be split into 3t2 sin (t3) and a stand-alone t.

Had the expression been, say, sin (t5), you would be looking for the chain "5t4".

Another way to do this question (it's actually easier this , so just for your learning since the hint specifies to integrate in another way ) :

Integrate cos(t³) by parts.

dv/dt - cos(t³) = -3t³sin(t³)
dv/dt = cos(t³) - 3t³sin(t³)


Integrate both sides with respect to t,

v = ∫cos(t³) dt - ∫3t³sin(t³) dt

Since cos(t³) = (1)(cos(t³)) , Let u = cos(t³) and let dv/dt = 1

(Formula : ∫(u dv/dt) dt = uv - ∫(v du/dt) dt)


v = cos(t³)∫(1)dt - ∫( ∫(1)dt (-3t²sin(t³)) ) dt
- ∫ 3t³sin(t³) dt

v = tcos(t³) - ∫ (-3t³sin(t³)) dt - ∫ 3t³sin(t³) dt

v = tcos(t³) + ∫ (-3t³sin(t³)) dt - ∫ 3t³sin(t³) dt

v = tcos(t³) + C , C is a constant


When v(0) = 0,

0 = 0(cos 0³) + C
C = 0

So v = tcos(t³)


Always check if integrals can be eliminated.

In addition, try to manipulate the equation and see if you can arrange the terms such that you obtain derivatives (of product rule, quotient rule, chain rule type etc)

If possible, you can simply get back the original term by integration